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How can I show that $$ \sigma_\phi \circ \sigma_\psi=\rho_{2(\phi-\psi)}, $$ where $\sigma_\phi$ is a reflection about a line making an angle $\phi$ with the x-axis, and $\rho_\psi$ is a rotation about the origin with angle $\psi$. Is it possible to show this is true geometrically? Or is the only way to do this multiply the matrices that correspond with the rotation and reflection?

EDIT

The way I see it now is:

Consider a vector on the x-axis (wlog), the vector first moves $2\psi$, and then $2(\phi-2\psi)$. So we get: $2\psi+2(\phi-2\psi)=2(\phi-\psi).$

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    Matrix Multiplication will be the quick, sure fire way to see this ... there probably is a neat geometric demonstration ... ask if you want the matrix calculation. – 2017-02-20

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You said in the chat that you wanted a geometric interpretation of this problem, so here it is! Consider the point $X$ we're applying these two reflections on. Consider $X' = \sigma_\psi(X)$ and $X'' = \sigma_\phi(X') = (\sigma_\phi \circ \sigma_\psi)(X)$.

Let's draw a figure. Now, notice that line $X'X$ is perpendicular to the first line of reflection. Also, $X'X''$ is perpendicular to the second line of reflection.

Perpendicular

Therefore, we conclude that $OX = OX' = OX''$; Therefore, there is a circle centered at $O$ through these three points:

enter image description here

Now what is $\angle XOX''$? Well, since rotation preserves angles, let's WLOG assume $X$ lies on the $x$-axis.

enter image description here

Now, $\angle X''OX = \angle X''X' + \angle X'X = 2\psi + 2\phi$. Therefore, the angle counterclockwise is $-2\psi + 2\phi = 2(\phi-\psi)$.

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    Thanks for pointing out the circle! The drawing was a bit confusing at first - and I must admit that I skipped your answer first - but it is the best drawing once you get the hang of it. – 2017-02-21
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